A proton with an initial speed of 8.20×105 m/s is brought to rest by an electric field. Part APart complete Did the proton move
1 answer:
Answer:
(a) Region of higher potential
(b) ΔV=1.911V
Explanation:
For Part (a)
Proton moves into a region of higher potential because it must goes in opposite direction of force to get rest
For Part (b)
Potential difference we can calculate bu definition of kinetic energy
ΔK=mv²/2= -qΔV
express in ΔV
ΔV= -mv²/2q
Substitute the given values
So
ΔV= -mv²/2q
ΔV=1.911V
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Answer:
Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives
For the y-direction gives
Combining both equation yields the y_component of the final velocity
Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.